Theorem ddif 4089

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif	⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Proof of Theorem ddif

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		velcomp 3914	. . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴)
2	1	con2bii 359	. . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴))
3		vex 3452	. . . 4 ⊢ 𝑥 ∈ V
4	3	biantrur 537	. . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)))
5	2, 4	bitr2i 278	. 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴)
6	5	difeqri 4077	1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴

Syntax hints:  ¬ wn 3   ∧ wa 398   = wceq 1554   ∈ wcel 2136  Vcvv 3448   ∖ cdif 3896

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-8 2138  ax-9 2146  ax-ext 2728

This theorem depends on definitions:  df-bi 209  df-an 399  df-tru 1557  df-ex 1794  df-sb 2085  df-clab 2735  df-cleq 2748  df-clel 2831  df-v 3450  df-dif 3902

This theorem is referenced by:  complss  4099  dfun3  4223  dfin3  4224  invdif  4226  ssindif0  4412  difdifdir  4439